Answer: 343g/mol
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=K_f\times m[/tex]
[tex]\Delta T_f=i\times K_f\times m\\\\\Delta T_f=i\times K_f\times\frac{\text{Mass of syucrose}}{\text{Molar mass of sucrose}\times \text{Mass of water in Kg}}[/tex]
[tex]\Delta T_f=T_f^0-T_f=(0-(-0.082)^0C=0.082^0C[/tex] = Depression in freezing point
[tex]K_f[/tex] = freezing point constant = [tex]1.86^0C/kgmol[/tex]
m= molality
i = Van't Hoff factor = 1 (for non electrolyte)
Now put all the given values in this formula, we get
[tex]0.082^0C=1\times (1.86^0C/kgmole)\times \frac{12.1g}{\text{Molar mass of sucrose}\times 0.8kg}[/tex]
[tex]\text{Molar mass of sucrose}=343g/mol[/tex]
Thus the molar mass of sucrose is 343 g/mol
